If P & Q are the ends of a pair of conjugate diameters & C is the centre of the ellipse `4x^2+9y^2=36`. Then the area of `triangleCPQ` is:

Answer

To solve the problem of finding the area of triangle \( CPQ \) where \( P \) and \( Q \) are the ends of a pair of conjugate diameters of the ellipse given by the equation \( 4x^2 + 9y^2 = 36 \), we can follow these steps:

### Step 1: Rewrite the equation of the ellipse
The given equation of the ellipse is:
\[ 4x^2 + 9y^2 = 36 \]
To standardize this equation, we divide each term by 36:
\[ \frac{4x^2}{36} + \frac{9y^2}{36} = 1 \]
This simplifies to:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]
From this, we can identify \( a^2 = 9 \) and \( b^2 = 4 \), which gives us \( a = 3 \) and \( b = 2 \).

### Step 2: Identify the center of the ellipse
The center \( C \) of the ellipse is at the origin:
\[ C(0, 0) \]

### Step 3: Determine the lengths of the semi-major and semi-minor axes
From the standard form of the ellipse, we have:
- Semi-major axis \( a = 3 \) (along the x-axis)
- Semi-minor axis \( b = 2 \) (along the y-axis)

### Step 4: Find the coordinates of points \( P \) and \( Q \)
Since \( P \) and \( Q \) are the ends of conjugate diameters, we can denote their coordinates as:
- \( P(a \cos \theta, b \sin \theta) = (3 \cos \theta, 2 \sin \theta) \)
- \( Q(a \cos(\theta + 90^\circ), b \sin(\theta + 90^\circ)) = (3 \cos(\theta + 90^\circ), 2 \sin(\theta + 90^\circ)) = (-2 \sin \theta, 3 \cos \theta) \)

### Step 5: Calculate the area of triangle \( CPQ \)
The area \( A \) of triangle \( CPQ \) can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, we can take \( CQ \) as the base and \( CP \) as the height.

1. The length of \( CQ \):
\[ CQ = \sqrt{(-2 \sin \theta - 0)^2 + (3 \cos \theta - 0)^2} = \sqrt{(2 \sin \theta)^2 + (3 \cos \theta)^2} = \sqrt{4 \sin^2 \theta + 9 \cos^2 \theta} \]

2. The length of \( CP \):
\[ CP = \sqrt{(3 \cos \theta - 0)^2 + (2 \sin \theta - 0)^2} = \sqrt{(3 \cos \theta)^2 + (2 \sin \theta)^2} = \sqrt{9 \cos^2 \theta + 4 \sin^2 \theta} \]

### Step 6: Use the area formula
Using the coordinates of points \( P \) and \( Q \):
\[ A = \frac{1}{2} \times CQ \times CP \]
However, we can simplify this further by recognizing that the area of triangle \( CPQ \) formed by the conjugate diameters can also be expressed as:
\[ A = \frac{1}{2} \times a \times b = \frac{1}{2} \times 3 \times 2 = 3 \]

### Final Answer
Thus, the area of triangle \( CPQ \) is:
\[ \boxed{3} \text{ square units} \]